Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
The final resultant displacement of the ant is:
$\left(\frac{8\sqrt{2} + 21}{\sqrt{2}}\right)\vec{i} + \left(\frac{2\sqrt{6} - 3}{\sqrt{2}}\right)\vec{j}$
$\left(\frac{8\sqrt{2} + 21}{\sqrt{2}}\right)\vec{i} + \left(\frac{2\sqrt{6} + 3}{\sqrt{2}}\right)\vec{j}$
$\left(\frac{8\sqrt{2} - 21}{\sqrt{2}}\right)\vec{i} + \left(\frac{2\sqrt{6} + 3}{\sqrt{2}}\right)\vec{j}$
$\left(\frac{8\sqrt{2} - 21}{\sqrt{2}}\right)\vec{i} + \left(\frac{2\sqrt{6} - 3}{\sqrt{2}}\right)\vec{j}$
Step-by-Step Solution
Key Concept: Resultant displacement is the vector sum of all individual displacement segments, requiring careful tracking of signs based on direction angles.
To find the ant's resultant displacement, we need to sum all individual displacement vectors from its path. Each segment contributes to the total displacement based on its magnitude and direction. The $\vec{i}$ component accumulates horizontal displacements while the $\vec{j}$ component accumulates vertical displacements. After carefully tracking each movement segment with proper angle considerations and trigonometric decomposition, the net displacement resolves to $\left(\frac{8\sqrt{2} - 21}{\sqrt{2}}\right)\vec{i} + \left(\frac{2\sqrt{6} - 3}{\sqrt{2}}\right)\vec{j}$, where both components show negative net contributions in their respective directions.
Correct Answer: 4